# Continuous variable depending on a Poisson process.

Let $$\ A,B$$ be a poisson process with $$\ \lambda = 2$$ in a timeframe of $$\ 1$$ minute. Both are independent variables. Let $$\ T$$ = time that has passed from $$\ 0$$ until the occurrence of the first event of $$\ A$$

I need to compute $$\ P(T > 0.25)$$

According to the solution given, $$\ T \sim exp(2)$$ and I can not understand why.

So I understand that time between two events of a poisson process is exponentially distributed yet I can't understand how to compute the parameter? In this question it means I have a event every 30 seconds. So if I understand correctly the time between events has a mean of $$\ 30$$ seconds so $$\ T \sim exp(30)$$ ?

Note $$T>t$$ if and only if there are no occurrences in the interval $$[0,t]$$. The number of occurrences in $$[0,t]$$ has a $$\text{Poi}(2t)$$ distribution. Therefore, $$P(T>t)=P(\text{Poi}(2t)=0)=e^{-{2t}}\quad\implies\quad P(T\le t)=1-e^{-2t}$$ Note that this is exactly the cdf of an $$\text{Exp}(2)$$ distribution.
Here is why $$T$$ is not distributed like $$\text{Exp}(30\text{ sec})$$. The parameter $$\lambda$$ is not the mean of $$T$$; the mean of $$T$$ is $$1/\lambda$$. Since, as you intuited, $$T$$ has a mean of $$30\text{ sec}=1/2\text{ min}$$, it follows, that $$E[T]=1/\lambda=1/2$$, so $$\lambda=2$$. Think of $$\lambda$$ as a "rate" parameter. $$T$$ is a wait time, and the larger $$\lambda$$ is, the faster that this time passes.